Abaseball is thrown horizontally off a cliff with a speed of 20 m/s. what is the horizontal distance, to the nearest tenth of a meter, from the face of the cliff after 4.9 seconds? 8 in the previous problem, to the nearest tenth of a meter, how far has it fallen in that time?

a) 98 m

b) 117.8 m

Explanation:

Hi!

Since there is no force acting on the horizontal direction, the horizontal displacement x as a function of time t is:

x(t) = 20 m/s * t

Therefore, the horizontal distance after 4.9 seconds is:

x(4.9 s) = (20 m/s) * (4.9 s) = 98 m

On the other hand, on the vertical direction, there is no initial velocity, but a constant acceleration g is acting on this direction, therefore, the vertical displacement is given by:

y(t) = - (1/2)*g*t^2

Therefore, the vertical distance after 4.9 seconds is:

y(4.9 s) = - (1/2) * (9.81 m/s^2)*(4.9 s) ^2 = - 117.769 m = -117.8 m

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio

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